Let $S$ be a Burniat surface with $K^2_S = 6$. Then we show that $\operatorname{glct}(S, K_S) = \frac{1}{2}$ by showing that $\operatorname{glct}(S, 2K_S) = \operatorname{lct} (S, E) = \frac{1}{4}$ for some divisor $E \in \lvert 2K_S \rvert$. This implies that Tian’s conjecture (which fails in general) holds for the polarized pair $(S, 2K_S)$, since the corresponding graded algebra is generated by sections of $H^0 (S, 2K_S)$. Moreover we verify that any divisor $D \in \lvert mK_S \rvert$ such that $\operatorname{glct}(S, K_S) = \operatorname{lct}(S, \frac{1}{m} D)$ for a positive even integer $m$ is invariant under the $\mathbb{Z}^2_2$-action associated to the bicanonical map of $S$.

中文翻译：

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$ K ^ 2 = 6 $的Burniat曲面的对数正则阈值

假设$ S $是Burniat曲面，其中$ K ^ 2_S = 6 $。然后我们通过显示$ \ operatorname {glct}（S，2K_S）= \ operatorname {lct}（S，E）来显示$ \ operatorname {glct}（S，K_S）= \ frac {1} {2} $ = \ frac {1} {4} $表示除数$ E \ in \ lvert 2K_S \ rvert $。这意味着Tian的猜想（通常会失败）适用于极化对$（S，2K_S）$，因为相应的渐变代数是由$ H ^ 0（S，2K_S）$的部分生成的。此外，我们验证\ lvert mK_S \ rvert $中的任何除数$ D \ d使得$ \ operatorname {glct}（S，K_S）= \ operatorname {lct}（S，\ frac {1} {m} D）$在与$ S $的经典映射相关的$ \ mathbb {Z} ^ 2_2 $作用下，一个正偶整数$ m $是不变的。